You need to explain it better. But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through. Korekt me if ayem rong, but I do not b-leave aye am.

You need to explain it better. But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through. Korekt me if ayem rong, but I do not b-leave aye am.

You need to explain it better. But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through. Korekt me if ayem rong, but I do not b-leave aye am.

You need to explain it better. But in a static system with a constant force (in this case weight) on one end of the rope, and a constant (in this case the anchor) opposing force on the other end of the rope, the tension will be the same no matter what angles it moves through. Korekt me if ayem rong, but I do not b-leave aye am.

I was just thinking "should I reply, or is it going too far?" Probably the latter... Anyways, yeah, explain the Q farther unless my answer that I answered answered the question that you questioned us with.

If you are belaying off of the anchor then the tension in the rope and the force on the anchor is equal to the force exerted by the person at the end.

[edited to change:] With a redirect belay (and no friction) the tension is equal to the force exerted by the belayer plus that of the climber which is approximately equal to the force on the anchor.

With a redirect belay the tension on the rope is still the same since the climber is not accelerated but the belayer end of the rope has the same tension as well. This puts twice the force on the anchor. [/edit]

Geometry and friction modify this force. I think biners are like 60% efficient as pulleys so it only requires about 60% of the force of the climber to hold her.

I hope that's what you were asking. Majid brings this up all the time so I'm sure he's proud of you for asking.

Are you sure the tension increases? Maybe I'm getting confused, it is late, but I'm thinking the force on the rope can't change, as then the climber would be pulled up, no? The force on the anchor would... aw naw! now I'm as confused as the OP!

When toproping and the climber falls and you get lifted off the ground you are both hanging from the rope.

When belaying off the anchor only one person is hanging from the rope.

You do the math.

That’s poor logic. In fact a fall straight onto an anchor produces more force on the anchor then when the rope runs through the anchor to the belayer. When you fall the belayer is lifted off the ground which increases the fall distance and decreases the peak load on the quickdraw / cam / anchor.

Belaying straight off the anchor for a second will produce more force on the anchor then doing a redirectional into your harness.

No, he's about right... What USnavy is saying is bad logic. You are assuming slack is developed in the belay, causing a fall before the rope gets taught. If a top belay is kept tight, it will experience HALF as much force as a toprope, or slingshot belay.

Add any slack, and the dynamics of the belay come into play, and the forces change. It's much harder to give a dynamic belay off the anchor, especially with an auto-locking device, so it is very important to keep any slack out of the rope.

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

I can't believe I'm going to say this, but this is where Majid's diagrams help to make it simple. Force on the anchor from the climber falling is easy to grok, so we'll just think about the other forces that can come into play.

"is tension on a rope doubled in a top rope system when its on a redirect belay?"

Sort of. Replace "tension on the rope" with "force on the anchor". Now a portion of the belayers body weight is also added to the downward force, but only to the extent required to counter-balance the falling climber. How much force that is can be tricky to nail down due to variables like the friction and angles in the redirect/slingshot belay system. 60% seems like a reasonable estimate, but is far from a reliable constant. Is the rope new and slick or all beat to shit and fuzzy like mine? Is the blayer right under the anchor or off to the side some?

"is tension on a rope doubled in a top rope system when its ... direct from the anchor belay."

No. Because the belay device attached to to the anchor catches the falling climber and holds only the force of the falling climber.

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

jt512 has given a clear and correct explanation. USNavy, I'm afraid, is misreading his bibles.

For those who are puzzled by the rope tension going up to a (momentary) peak load of twice the climber's weight, remember that there is still rope stretch when a climber just hangs on the rope, the moment of maximum stretch produces the peak tension in the rope, and that peak tension turns out to be twice the weight hung on the rope.

A redirected top-rope anchor has to be able to withstand far more than triple body weight to be even remotely safe, since most belayed falls with have small but non-zero fall factors.

A redirected top-rope anchor has to be able to withstand far more than triple body weight to be even remotely safe, since most belayed falls with have small but non-zero fall factors.

Which leads to the question: How strong should a toprope anchor be to be more than just remotely safe?

We saw, above, that when a 200-lb climber takes a toprope fall, the peak force on the anchor will be at least 10/3 times his weight, or 667 lb. Since some climbers weigh more than 200 lb, and most toprope falls will have fall factors greater than zero, let's round this number up to 1000 lb. Now multiply by a safety factor of 3, and I claim that a toprope anchor should be built to withstand a force of at least 3000 lb.

is tension on a rope doubled in a top rope system when its on a redirect belay or a direct from the anchor belay.

it seems like it should but im draawing a blank if not please say why

I have read through the responses listed above and am confused. I must not understand the question. What are you calling a "redirect belay"? Is that the belayer on the ground where the climber began and the rope through a single overhead anchor?

If you tie the rope at the anchor and hang from it. It has your weight as tension. If you double the rope through the anchor and it is able to slide. Then hang from 1 end you will fall. So, you tie off the other side to counter your weight. The rope has the same tension as before. You just moved the tie off place and have twice as much rope in play.

This does assume no movement. Just a static situation. Falling on the rope and movement of the rope complicates this.

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight — that is, a little more than triple the climber's weight.

Jay

where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

edited to add: if this is all explained in the above link to Rich's paper, then never mind.